Solutions MCQs for NEET — Chemistry Questions with Answers

$$\begin{gathered} 1\mathrm{N} {\mathrm{KNO}}_3=1\mathrm{M} {\mathrm{KNO}}_3 = 1 \times 2\mathrm{M} (\mathrm{Effective} \mathrm{Molarity}) \\ 1\mathrm{N} \mathrm{Ba}({\mathrm{NO}}_3{)}_2 = \frac{\mathrm{M}}{2}\mathrm{Ba}({\mathrm{NO}}_3{)}_2=\frac{1}{2}\times 3\mathrm{M} (\mathrm{Effective} \mathrm{Molarity}) \\ 1\mathrm{N} {\mathrm{Al}}_2({\mathrm{SO}}_4{)}_3 = \frac{\mathrm{M}}{6}{\mathrm{Al}}_2({\mathrm{SO}}_4{)}_3=\frac{1}{6}\times 5\mathrm{M} (\mathrm{Effective} \mathrm{Molarity}) \\ 1\mathrm{N} \mathrm{Th}({\mathrm{NO}}_3{)}_4 = \frac{\mathrm{M}}{4}\mathrm{Th}({\mathrm{NO}}_3{)}_4=\frac{1}{4}\times 5\mathrm{M} (\mathrm{Effective} \mathrm{Molarity}) \end{gathered}$$

Lesser is the molarity, higher is the vapour pressure. So, 1N Al₂(SO₄)₃ has highest vapour pressure.

K₄[Fe(CN)₆]$\to$4K⁺+[Fe(CN)₆]^4-

$\mathrm{i}=1+4\mathrm{\alpha }=1+4\times 0.6=3.4$

 $$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m} \\ 3.72=\mathrm{i}\times 1.86\times 1 \\ \mathrm{i}=2 \end{gathered}$$

273 -269

Hence complex must give 2 ions. So, complex is [Pd(H₂O)₃Cl₃]Cl.3H₂O

(d) For a dilute solution. the depression in freezing point (ΔT_f) is directly proportional to molality (m) of the solution:-

ΔT_f ∝ m or ΔT_f=K_fm

Where, K_f is called molal depression constant or freezing point depression constant or cryoscopic constant.The value of K_f ,depends only on nature of the solvent and independent of composition of solute particles, i.e. does not depend on the concentration of solution.

(b)Molarity and normality are temperature dependent because they involve volume of solutions.Volume is dependent on temperature.

Molarity(M)=Moles of solute/ Volume of solution (in L)

Molality,mole fraction and weight percentage do not depend on temperature because they involve masses of solute and solvent.

(d) Key idea Strong electrolytes dissociate completely in their solutions.

Van't Hoff factor=Total number of ions after dissociation
So,for Ba(OH)₂ â‡Œ Ba²⁺(aq)+2OH-(aq)
Number of ions           1            2
Van't Hoff factor, i=1+2=3

(d) Key Idea FOr this problem, the following expression can be used.

                      $∆{\mathrm{G}}_{\mathrm{mix}}=∆{\mathrm{H}}_{\mathrm{mix}}-\mathrm{T}∆{\mathrm{S}}_{\mathrm{mix}}$

For an ideal gas

                      $∆{\mathrm{H}}_{\mathrm{mix}}= 0; ∆{\mathrm{U}}_{\mathrm{mix}} =0; ∆{\mathrm{S}}_{\mathrm{mix}}\ne 0$

Putting all these values in the expression,

            $$\begin{gathered} ∆{\mathrm{G}}_{\mathrm{mix}}=∆{\mathrm{H}}_{\mathrm{mix}}-\mathrm{T}∆{\mathrm{S}}_{\mathrm{mix}} \\ ∆{\mathrm{G}}_{\mathrm{mix}}= 0-\mathrm{T}∆{\mathrm{S}}_{\mathrm{mix}} \\ \therefore ∆{\mathrm{G}}_{\mathrm{mix}} \ne 0 \end{gathered}$$

Thus, option (d) is incorrect.

(d) From Raoult's law of  paratial pressure,

(p_A-p_s)/p_s = n_B / n_A

= 760-732 / 732 = W_B x M_A / M_B x W_A

=28/732 = 6.05x18 /M_A x 100

= M_A = 30.6

$∆$Tb = 0.52x6.5x1000/30.6x100 = 1.10

Boiling point = 100 + 1.10

=101.1°C=101°C

(d) Since, component having higher vapour pressure will have higher percentage in vapour phase. Benzene has vapour pressure 12.8 kPa which is greater than toluene 3.85 kPa.

Therefore, the vapour will contain a higher percentage of benzene.

For an ideal solution:-
(i)There will be no change in volume on mixing the two components i.e. ΔV_mixing = 0.
(ii)There will be no change in volume on ΔH_mixing = 0. So, ΔS_mix≠0 for an ideal solution.